Conditional Probability: John and Mary's registration cards

Question

I am trying to solve a problem from a Dartmouth textbook (https://math.dartmouth.edu/~prob/prob/prob.pdf) Chapter 4.1 Ex 53.

Disclaimer: I am not a student. I am just studying some probability questions from the text book.

Here is the question:

The registrar is carrying John and Mary’s registration cards and drops them in a puddle. When he pickes them up he cannot read the names but on the first card he picked up he can make out Mathematics 23 and Government 35, and on the second card he can make out only Mathematics 23. He asks you if you can help him decide which card belongs to Mary. You know that Mary likes government but does not like mathematics. You know nothing about John and assume that he is just a typical Dartmouth student. From this you estimate:

P (Mary takes Government 35) = .5

P (Mary takes Mathematics 23) = .1

P (John takes Government 35) = .3

P (John takes Mathematics 23) = .2

Assume that their choices for courses are independent events. Show that the card with Mathematics 23 and Government 35 showing is more likely to be Mary’s than John’s. The conjunction fallacy referred to in the Linda problem would be to assume that the event “Mary takes Mathematics 23 and Government 35” is more likely than the event “Mary takes Mathematics 23.” Why are we not making this fallacy here?

Here is the solution from the manual:

We assume that John and Mary sign up for two courses. Their cards are dropped, one of the cards gets stepped on, and only one course can be read on this card. Call card I the card that was not stepped on and on which the registrar can read government 35 and mathematics 23; call card II the card that was stepped on and on which he can just read mathematics 23. There are four possibilities for these two cards. They are:

Card I          Card II          Prob.  Cond. Prob.
Mary(gov,math)  John(gov, math)  .0015  .224
Mary(gov,math)  John(other,math) .0025  .373
John(gov,math)  Mary(gov,math)   .0015  .224
John(gov,math)  Mary(other,math) .0012  .179

In the third column we have written the probability that each case will occur. For example, for the first one we compute the probability that the students will take the appropriate courses: .5 × .1 × .3 × .2 = .0030 and then we multiply by 1/2, the probability that it was John’s card that was stepped on. Now to get the conditional probabilities we must renormalize these probabilities so that they add up to one. In this way we obtain the results in the last column. From this we see that the probability that card I is Mary’s is .597 and that card I is John’s is .403, so it is more likely that that the card on which the registrar sees Mathematics 23 and Government 35 is Mary’s.

My question:

It seems like the solution calculates the Probability of "Other" by

Prob(John takes other) = 1 - Prob(John takes gov) - Prob(John takes math) = 1 - 0.3 - 0.2 = 0.5

See row 2: Mary(gov,math) John(other,math), this is calculated as 0.5*0.1*0.5*0.2 / 2 = 0.0025

This subtraction above implies "John takes gov" is exclusive from "John takes math". But later they are multiplying Prob(John takes gov)*Prob(John takes math) to get the joint probability. I find this problem/solution not making sense.

What do you guys think? Alternatively, how would you best understand the solution to this question to make the most sense out of it?

Answer 1

You're right that the solution is inconsistent. Already the first sentence in the solution is inconsistent with the problem statement: If the choices for courses are independent events, as stipulated in the problem statement, then students cannot have a fixed number of $2$ courses each. The only way I can see to make sense of this is to assume that the problem statement means only that the choices for/against math and government are independent, whereas the choices for/against other courses can be taken dependent on the math/government choices so as to reach a fixed total of $2$ courses. That would mean that the probability for John’s card to be (other,math) would be $(1-0.3)\cdot0.2$, the probability that he didn't take government times the probability that he took math. As you say, the solution seems to use the value $0.5$ instead of $1-0.3$. I see no valid interpretation that would lead to this value. As you say, it seems that in both cases of cards with another course, the probability for choosing that course has been calculated as the complement of the two probabilities for choosing the other two courses. If so, this is simply wrong.

Answer 2

I think your objection is well-taken. I thought perhaps that we could rescue the given solution by saying that the student can take two math classes, and the probability of taking math in the second course is the same as taking math in the first course. But then shouldn't the probability that John takes math be $1-.8^2=.36$?

I can't find a way to to make sense of the given solution.

Answer 3

It seems that the given solution is completely wrong, for the reasons you have detailed. Not only that, but the whole question is quite a mess — the author mentions the conjunction fallacy and conditional probabilities, but the fallacy is not noteworthy here (not that I can see anyway), and conditional probabilities are not useful in solving this problem (in my opinion; see my final point). However, the problem is still tractable (assuming my working is correct). When I worked through the problem, I used Venn diagrams to force me to evaluate all of the probabilities. It was largely pointless, since the majority of the probabilities I evaluated went unused in the calculations, but it meant that I had all of the probabilities for all of the events on the page in front of me when I needed them. That is to say, using Venn diagrams may help you tackle problems like this. I've omitted all that needless working here, as it boils down to the following...

We have two mutually exclusive cases to consider:

1. Card I belongs to Mary (so Mary takes both classes), and Card II (which the registrar stepped on) belongs to John (so John takes Mathematics).
2. Card I belongs to John (so John takes both classes), and Card II (which the registrar stepped on) belongs to Mary (so Mary takes Mathematics).

Let $S_C$ denote the event "student $S$ takes class $C$", and let $S_\text{trodden}$ denote the event "the registrar stepped on the card belonging to student $S$". Let $M$ denote Mary and $J$ denote John.

Then our cases correspond to the following events:

1. $\big( M_\text{Gov} \cap M_\text{Math} \big) \cap J_\text{trodden} \cap J_\text{Math}$
2. $\big( J_\text{Gov} \cap J_\text{Math} \big) \cap M_\text{trodden} \cap M_\text{Math}$

Assume that their choices for courses are independent events.

Using this fact, we can determine the probabilities of the parenthesised events:

1. $\mathrm{P} \big( M_\text{Gov} \cap M_\text{Math} \big) = \mathrm{P} \big( M_\text{Gov} \big) \times \mathrm{P} \big( M_\text{Math} \big) = 0.5 \times 0.1 = 0.05$
2. $\mathrm{P} \big( J_\text{Gov} \cap J_\text{Math} \big) = \mathrm{P} \big( J_\text{Gov} \big) \times \mathrm{P} \big( J_\text{Math} \big) = 0.3 \times 0.2 = 0.06$

We assume that the registrar would step on either card with equal likelihood, i.e.

$$\mathrm{P}\big( J_\text{trodden} \big)= \mathrm{P}\big( M_\text{trodden} \big) = 0.5\text{.}$$

We already know the probabilities of the events $J_\text{Math}$ and $M_\text{Math}$; they're $0.2$ and $0.1$, respectively.

Thus, the probabilities of our cases, respectively, are:

1. $0.05 \times 0.5 \times 0.2 = 0.005\text{,}$
2. $0.06 \times 0.5 \times 0.1 = 0.003\text{.}$

Hence, Case 1 is more likely; so the textbook author's end result is correct, but his reasoning and calculated probabilities are wrong.

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Where conditional probabilities might be considered useful is in observing that the info we have implies that both Mary and John are taking the Mathematics class. Thus, because each choice of which class to take was made independently, we can "factor out"/ignore those events when looking at our cases. That is, we can instead compare the probabilities of the following events, respectively, to determine the answer:

1. $M_\text{Gov} \cap J_\text{trodden}$, which has probability $0.5 \times 0.5 = 0.25$.
2. $J_\text{Gov} \cap M_\text{trodden}$, which has probability $0.3 \times 0.5 = 0.15$.

Intuitively, then, the result seems obvious in hindsight; the registrar is equally likely to step on either card, and both students are known to be taking the Mathematics class, so all that one needs to consider is how likely each student is to be taking the Government class. Whichever student is more likely to do so is more likely to have Card I belong to them.